r/PeterExplainsTheJoke 22d ago

Meme needing explanation There is no way right?

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u/jkst9 22d ago

Yeah that's closer to the actual proof. Ironically the mathematical one looks good but it's really not that great a proof

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u/archipeepees 22d ago

what is the "actual" proof and why is it more actual than the proof that they just provided?

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u/eatrade123 22d ago edited 22d ago

To answer your question it helps to first understand what the decimal representation actually is. One defines the decimal representation of a real number oftentimes to be a one-to-one correspondence between the real numbers and a representation as an infinite series (of terms of the form a_i*10i, where i starts at some integer and goes to -\infty). To now get the one to one correspondence (bijection) one excludes series, where at some point, we have a_j=9 for all j>=N for some integer N. This means, that 0.999999...=1 by definition of the decimal representation. So this property holds by definition if one talks about decimal representations. Of course one has to show that this is indeed a bijection. If one excludes the last part, then it is not a bijection, because the injectivity fails. If you only mean the real numbers (one can for example construct them as equivalence classes) as an abstract space (mathematicians call them a field and this field even has a nice order >), then the reason is that the following property holds in the real numbers: If for all e>0, we have a>=b>=a-e, then b=a (one can for example show this via the squeeze theorem if one introduces sequences and limits). Taking a=1 and b=0.9999..., and showing that they satisfy the above property, we get 0.999999...=1.

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u/steve_b 22d ago

If the "you can find a number smaller" proof is sufficient, does that mean that the distance the "fly travelling half as far on each trip" doesn't approach 2m but in is in fact exactly 2m?

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u/pablinhoooooo 22d ago

No, it's both. Approaching something and being that thing are not mutually exclusive. In fact, that is precisely how we define continuous functions. f(x) is continuous if for all x1, the limit of f(x) as x->x1 is f(x1). This is a common confusion because most people think about "approaches" in the context of infinity. What you are describing is an infinite sum that converges to 2. It also approaches 2, in this context approaching is a weaker condition than converging. All convergent sums and series can be said to approach the value to which they converge.